Factor completely. $8 x^2 +112 x +392=$
First, we take a common factor of $8$. $8 x^2 +112 x +392=8(x^2 +14 x +49)$ Now, let's factor $x^2 +14 x +49$. Both $x^2$ and $49$ are perfect squares, since $x^2=({x})^2$ and $49=({7})^2$. Additionally, $14 x$ is twice the product of the roots of $x^2$ and $49$, since $14 x=2({x})({7})$. $x^2 +14 x +49 = ({x})^2 + 2({x})({7})+({7})^2$ So we can use the square of a sum pattern to factor: ${a}^2 + 2( a)( b)+ {b}^2 =({a} + {b})^2$ In this case, ${a}={x}$ and ${b}={7}$ : $ ({x})^2 + 2({x})({7})+({7})^2 =({x} +{7})^2$ $\begin{aligned} 8 x^2 +112 x +392 &=8(x^2 +14 x +49) \\\\ &=8(x +7)^2 \end{aligned}$ In conclusion, the complete factorization is $8(x +7)^2$ Remember that you can always check your factorization by expanding it.